Conference rusure::math

[2] = [3], 

[4] = [5]
      

/Bevin

I'm afraid [2] <> [3] and [4] <> [5].  My guess is:

	[3] > [2] > [1] > [4] > [5].


				-- edp

My guess would be [3] > [2] > [1] > [5] > [4]

  John

Why should the fact that it is known to be the ace OF SPADES affect the
probability, compared to just knowing it is some ace?

The argument is simple.  Pull an ace, any ace, out of the deck.  What is
the probability that in drawing the remaining cards you will pull another
ace?  Obviously this just depends on the fact there are 51 cards left, including
3 aces.

Now, pull the ace of SPADES out of the deck.  What is the probability that
...  Obviously this just depends ...

The calculations are identical, hence will lead to the same result.

Hence [2] = [3], and mutatis mutandis (sp?) [4] = [5]

/Bevin

Could someone a supply a simple argument for why [3] > [2]?

Re .5:

Here is a demonstration that may help intuitively.  Figures for a real deck
follow this.

Consider a smaller deck, consisting of just the ace of spades (AS), the ace of
clubs (AC), and the seven of hearts (7H).  Let a hand consist of just two
cards.  Then all of the hands are:

	AS AC,
	AC 7H, and
	7H AS.

Clearly, the probability that exactly two cards are aces is 1/3.

Suppose someone tells us one of the cards is an ace.  We still have the same
set of possible hands, so the probability that exactly two cards are aces is
1/3.

Suppose someone tells us one of the cards is the ace of spades.  Then the
possible hands are:

	AS AC and
	7H AS.

Thus, the probability that exactly two cards are aces is now 1/2.

In what follows (a b) denotes the number of combinations of a things taken
b at a time.  So (5 2) = 5! / (3! * 2!) = 10.

To see this with a complete deck and regular bridge hands, observe that
(4 2)(48 11) is the number of hands with exactly two aces (two cards are
selected from four aces; eleven cards are selected from the other 48 cards).
The number of hands with at least one ace is

	(52 13) - (4 0)(48 13).

That is the total number of hands minus the number of hands with no aces.
So the probability that exactly two cards are aces when at least one card is
an ace is:

	(4 2)(48 11) / [(52 13) - (4 0)(48 13)] = 2223/7249 ~ .3067.

(~ denotes "approximately".  Somebody please check this, I haven't.)

On the other hand, the number of hands with exactly two aces, one of which is
the ace of spades, is (1 1)(3 1)(48 11).  That is, one card is chosen from the
set containing only the ace of spades, one card is chosen from the other aces,
and eleven cards are chosen from the rest of the cards.  The number of hands
containing the ace of spades is (1 1)(51 12).  This is choosing one card from
the set containing only the ace of spades and twelve cards from the rest of the
deck.  So the probability that exactly two cards are aces when one card is the
ace of spades is:

	(1 1)(3 1)(48 11) / (1 1)(51 12) = 8892/20825 ~ .4270.

(Please check this too.)

				-- edp

And where does the answer to [6] fit in with the other answers?

[6]	What is the probability that a bridge hand contains exactly two
	aces, if the top card is an ace?

Re .7:

What is the top card?  The last card dealt or the highest in value?

If the top card is the highest in value, then

	The top card is an ace if and only if there is at least one
	ace in the hand.

Thus, [6] = [2].


				-- edp

The question in .7 was intended to refer to the last card dealt, and the
note was intended to relate to .5

    re .6
    Not so with the three card deck example. 
    AS AC
    AS 7H       The chance of having two Aces is 1/3 .  Knowing that
    7H AC       the AS is held the chance of two Aces is 1/2 .  Knowing
                that one of the cards is an Ace, the chance of having
    two Aces is still 1/2 .  Each Ace is unique so the hand with the
    other Ace does not become one of the possibilities (1/3).  If you
    know the AC is held the chance of two Aces is still 1/2 .  Knowing
    the hand holds an Ace makes it either an AS or an AC but not both.

    Re .10:
    
    That is not correct.  A simple experiment will demonstrate it. 
    If you wish, I will take a deck consisting of the ace of spades,
    the ace of clubs, and the seven of hearts, and I will randomly select
    hands of two cards from it.  (Since the deck is too small to shuffle,
    I will use a die to select the hand.)  Each time I select a hand,
    I will send you a mail message telling you there is an ace in the
    hand.  Since you believe there is a fifty-fifty chance that the
    hand consists of two aces, you should be willing to wager one dollar
    that the hand is two aces against my one dollar that the hand is
    not two aces.  After enough such wagers, you will be broke.
    
    Observe the possibilities:
                                                                               
    	Hand is AC, AS.  I send you mail saying there is an ace.  You wager
    		the hand is two aces.  You win.  I pay you $1.
    	Hand is AC, 7H.  I send you mail saying there is an ace.  You wager
    		the hand is two aces.  You lose.  You pay me $1.
    	Hand is AS, 7H.  I send you mail saying there is an ace.  You wager
    		the hand is two aces.  You lose.  You pay me $1.
    
    Would you like to try it (even for pretend dollars)?
    
      
    				-- edp

    Re .11:
    
     My apologies to you.  After rereading the example you are entirely
    correct.  I believe I was thinking if the first card dealt in the
    hand was an Ace then obviously the chance of the second is 1/2,
    but we are looking at a hand after the deal.  Any Ace results in
    1/3 and a specific Ace results in 1/2.
    
    Where's my coffee!!  SN

	[this was done by Bill Long, an ex-Deccie]
	--------------------------------------------

	Aces in a bridge hand:

	1.	Total set of hands = (52 13) = 52!/39!13!

		Set of hands with two aces = (4 2)(48 11)

		Pr(two aces exactly) = (4 2)(48 11)/(52 13)

					= 4446/20825

					= 0.2135

	2.	Pr(exactly two aces, given hand includes at least
		   one ace):

		Pr(zero aces) = (4 0)(48 13)/(52 13) = 0.3038

		Pr(at least one ace) = 1-Pr(zero aces) = 0.6962

		no. of hands with zero aces = (4 0)(48 13)

		There is still an available set of (4 2)(48 11)
		hands with two aces, scattered in a set of all
		hands minus the aceless hands, given hand contains
		at least one ace.

		or:  Pr(2A/at least 1A) = 
				
			(4 2)(48 11)/[52!/39!13!-48!/35!13!]
					 ^	   ^
					 |	   |
			all hands -------	    ----- aceless hands

					= 2223/7249 = 0.3067

		The same result obtains if one simply divides 0.2135
		by 0.6962; that is the prob. of two aces is increased
		by setting aside the aceless hands (which comprise
		approximately three tenths of the total number of
		possible hands).



	3.	Pr(exactly two aces if hand is known to contain
		ace of spades):

		We know one fourth of all hands contain the ace of
		spades, or (52 13)/4.  To fill in the rest of the
		hand, we need to select one of the three remaining
		aces, plus eleven non-aces:  (3 1)(48 11)

		Pr(2A/Ace of spades) = (3 1)(48 11)/[(52 13)/4]

					= 8892/20825 = 0.4270

		We can write this result directly as:

			(1 1)(3 1)(48 11)/(51 12) = 0.4270

		Why is Pr(2A/Ace of spades) > Pr(2A/at least 1A)?

		We know that just half of the two-ace hands contain 
		the ace of spades.  Given that our hand contains the
		ace of spades, this tells us that HALF of the two-ace
		hands are located in the set of ONE-FOURTH of all
		possible hands.  On the other hand, given that our
		hand contains at least one ace, we have available ALL
		the exactly two-ace hands, contained in the set of
		0.6962 of the possible hands.  Considering the second
		case relative to the first, we are looking for twice
		as many two-ace hands, but in a set that is 0.6962/0.25
		times as large.  Thus:

	Pr(2A/Ace of spades) = 0.5(0.6962/0.25)xPr(2A/at least 1A)

		pr(2A/Ace of spades) = 1.3924xPr(2A/at least 1A)


	4. Pr(2A/at least 1K):

		hands with exactly 1K2A: (4 1)(4 2)(44 10) = 5.955e10
		 ..    ..    ..    2K2A: (4 2)(4 2)(44 9)  = 2.552e10
		 ..    ..    ..    3K2A: (4 3)(4 2)(44 8)  = 0.4254e10
		 ..    ..    ..    4K2A: (4 4)(4 2)(44 7)  = 0.0230e10

		All hands with at least 1K and exactly 2A = sum of
		above = 8.9954e10.

		The set of hands of interest is those containing at
		least 1K:
				= (52 13) - (4 0)(48 13)
				= 6.3501e11 - 1.9293e11
				= 4.4208e11 hands

		So Pr(2A/at least 1K) = 8.9954e10/4.4208e11
				      = 0.2035

		Another way to look at this:

		how many kingless two-ace hands are there?
			= (4 0)(4 2)(44 11) = 4.6016e10

		from (1), number of 2A hands = 1.3557e11; therefore,
		2A hands with at least 1K = 1.3557e11 - 4.6016e10

					  = 8.9554e10 (same as above)



	5.	Pr(exactly 2A/hand contains K spades):

		Hands containing K spades and exactly 2A:
					(1 1)(4 2)(47 10)

		Set of possible hands containing K spades:
					(1 1)(51 12)

		Therefore Pr(2A/K spades) = 6x[47!/37!10!][39!12!/51!]
					  = 8151/41650
					  = 0.1957

		Similar logic that explains why (3)>(2) applies in a
		reciprocal sense to explain why (4)>(5).